Question

differentiate the function. y=(4x - 4)^4(1 - x^3)^4
\frac{dy}{dx}=square

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. Let $u=(2x - 4)^4$ and $v=(1 - x^3)^4$.

Step2: Differentiate $u$ using chain - rule

If $u=(2x - 4)^4$, let $t = 2x-4$, then $u = t^4$. By the chain - rule $\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}$. $\frac{du}{dt}=4t^3$ and $\frac{dt}{dx}=2$. So $\frac{du}{dx}=4(2x - 4)^3\cdot2=8(2x - 4)^3$.

Step3: Differentiate $v$ using chain - rule

If $v=(1 - x^3)^4$, let $s = 1 - x^3$, then $v = s^4$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{ds}\cdot\frac{ds}{dx}$. $\frac{dv}{ds}=4s^3$ and $\frac{ds}{dx}=-3x^2$. So $\frac{dv}{dx}=4(1 - x^3)^3\cdot(-3x^2)=-12x^2(1 - x^3)^3$.

Step4: Calculate $y^\prime$

Using the product - rule $y^\prime=\frac{du}{dx}v+u\frac{dv}{dx}$, we substitute $u$, $v$, $\frac{du}{dx}$, and $\frac{dv}{dx}$:
\[

$$\begin{align*} y^\prime&=8(2x - 4)^3(1 - x^3)^4+(2x - 4)^4(-12x^2(1 - x^3)^3)\\ &=(2x - 4)^3(1 - x^3)^3[8(1 - x^3)-12x^2(2x - 4)]\\ &=(2x - 4)^3(1 - x^3)^3(8 - 8x^3-24x^3 + 48x^2)\\ &=(2x - 4)^3(1 - x^3)^3(8 + 48x^2-32x^3) \end{align*}$$

\]

Answer:

$(2x - 4)^3(1 - x^3)^3(8 + 48x^2-32x^3)$